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Where’s Bourbaki’s Escorial?
Feb 8th
Early 1936, Andre Weil and Evelyne Gillet made a pre-honeymooning trip to Spain and visited El Escorial. Weil was so taken by the place that he planned the next Bourbaki-conference to be held in a nearby college. However, the Bourbakis never made it to to Spain that summer as the Spanish Civil War broke out July 17th, a few weeks before the intended conference. Can we GEO-tag the exact location of Bourbaki’s “Escorial”?
As explained in the bumpy-road-post, Andre Weil and Evelyne Gillet became involved sometime in 1935. Early 1936, they made a pre-honeymooning trip to Spain and visited El Escorial. Weil was so taken by the place that he planned the next Bourbaki-conference to be held in a nearby college.
However, the Bourbakis never made it to to Spain that summer as the Spanish Civil War broke out July 17th, a few weeks before the intended conference. Still, the second Bourbaki-meeting remains often referred to as the ‘Escorial conference’. Can we GEO-tag the exact location of Bourbaki’s “Escorial”?
Claude Chevalley came up with a Plan-B and suggested they would use his parents’ place in Chançay as their venue. Chevalley’s father was a French diplomat and his house sure did possess a matching ‘grandeur’ as can be seen from the famous picture below, taken at the (second) Chançay meeting in 1937 (Weil to the left, Chevalley to the right and Weil’s sister Simonne standing).
Thanks to the Bourbaki archives we know that the meeting took place from september 16th to 28th, that each of them had to pay 16 francs for full pension and had to bring along their own sheets and towels.
But where exactly is this beautiful house? Jacques Borowczyk has written a nice paper Bourbaki et la touraine in which he describes the Bourbaki congresses of 1936 and 1937 at the Chevalley-house in Chançay and further those held in 1956, 1957 and 1959 in ‘hôtel de la Brèche’ in Amboise.
Borowczyk places the Chevalley house in the little hamlet of Chançay, called “La Massoterie”. The village files assert that in 1931 three people were living at La Massoterie : father Abel Chevalley, who took residence there after his retirement in 1931, his wife Marguerite and their son Claude. But, at the time of the Bourbaki congres in 1936, Marguerite remained the only permanent inhabitant. Sadly, Abel Chevalley, who together with Marguerite compiled the The concise Oxford French dictionary, died in 1934.
Usually when you know the name of the hamlet, of the village and add just to be certain ‘France’, Google Maps takes you there within metres. So, this was going to be a quick post, for a change… Well, much to my surprise, typing ‘La Massoterie, Chançay, France’ only produced the answer “We could not understand the location La Massoterie, Chançay, France”.
Did I spell it wrong? Or, did the name change over times? No, Googling for it the first hit gives you the map of a 10km walk around Chançay passing through la Massoterie!
Now what? Fortunately Borowczyk included in his paper an old map, from Napoleonic times, showing the exact location of La Massoterie (just above the flash-sign), facing the castle of Volmer. If you compare it with the picture below from present day Chançay (via Google earth) it is surprising how many of the landmarks have survived the changes over two centuries.
It is now easy to pinpoint the exact location and zoom into the Chavalley-house, and, you’re in for a small surprise : the place is called La Massotterie with 2 t’s…
Probably, Googles database is more reliable than the information provided by the village of Chançay, or the paper by Borowczyk as it is the same spelling as on the old Napoleonic map. Anyway, feel free to have a peek at Bourbaki’s Escorial yourself!
Lambda-rings for formula-phobics
Feb 5th
In 1956, Alexander Grothendieck (middle) introduced 
-rings in an algebraic-geometric context to be commutative rings A equipped with a bunch of operations 
(for all numbers 
) satisfying a list of rather obscure identities. From the easier ones, such as
to those expressing 
and 
via specific universal polynomials. An attempt to capture the essence of -rings without formulas?
Lenstra’s elegant construction of the 1-power series rings 
requires only one identity to remember
.
Still, one can use it to show the existence of ringmorphisms 
, for all numbers 
. Consider the formal ‘logarithmic derivative’
where 
is the usual formal derivative of a power series. As this derivative satisfies the chain rule, we have
and so all the maps are additive. To show that they are also multiplicative, it suffices by functoriality to verify this on the special 1-series 
for all 
. But,
That is, 
and Lenstra’s identity implies that 
is indeed multiplicative! A first attempt :
hassle-free definition 1 : a commutative ring 
is a -ring if and only if there is a ringmorphism 
splitting 
, that is, such that 
.
In particular, a -ring comes equipped with a multiplicative set of ring-endomorphisms 
satisfying 
. One can then define a -ringmorphism to be a ringmorphism commuting with these endo-morphisms.
The motivation being that -rings are known to form a subcategory of commutative rings for which the 1-power series functor is the right adjoint to the functor forgetting the -structure. In particular, if is a -ring, we have a ringmorphism 
corresponding to the identity morphism.
But then, what is the connection to the usual one involving all the operations ? Well, one ought to recover those from 
.
For 
to be a ringmorphism will require identities among the . I hope an expert will correct me on this one, but I’d guess we won’t yet obtain all identities required. By the very definition of an adjoint we must have that is a morphism of -rings, and, this would require defining a -ring structure on 
, that is a ringmorphism 
, the so called Artin-Hasse exponential, to which I’d like to return later.
For now, we can define a multiplicative set of ring-endomorphisms 
from requiring that 
for all . Another try?
hassle-free definition 2 : is a -ring if and only if there is splitting to satisfying the compatibility relations 
.
But even then, checking that a map is a ringmorphism is as hard as verifying the lists of identities among the . Fortunately, we get such a ringmorphism for free in the important case when A is of ‘characteristic zero’, that is, has no additive torsion. Then, a ringmorphism exists whenever we have a multiplicative set of ring endomorphisms 
for all such that for every prime number 
the morphism 
is a lift of the Frobenius, that is, 
.
Perhaps this captures the essence of -rings best (without the risk of getting an headache) : in characteristic zero, they are the (commutative) rings having a multiplicative set of endomorphisms, generated by lifts of the Frobenius maps.
Seating the first few thousand Knights
Feb 3rd
The Knight-seating problems asks for a consistent placing of n-th Knight at an odd root of unity, compatible with the two different realizations of the algebraic closure of the field with two elements. The first identifies the multiplicative group of its non-zero elements with the group of all odd complex roots of unity, under complex multiplication. The second uses Conway’s ‘simplicity rules’ to define an addition and multiplication on the set of all ordinal numbers.
The odd Knights of the round table-problem asks for a specific one-to-one correspondence between two realizations of ‘the’ algebraic closure 
of the field of two elements.
The first identifies the multiplicative group of its non-zero elements with the group of all odd complex roots of unity, under complex multiplication. The addition on is then recovered by inducing an involution on the odd roots, pairing the one corresponding to x to the one corresponding to x+1.
The second uses Conway’s ‘simplicity rules’ to define an addition and multiplication on the set of all ordinal numbers. Conway proves in ONAG that this becomes an algebraically closed field of characteristic two and that is the subfield of all ordinals smaller than 
. The finite ordinals (the natural numbers) form the quadratic closure of 
.
On the natural numbers the Conway-addition is binary addition without carrying and Conway-multiplication is defined by the properties that two different Fermat-powers 
multiply as they do in the natural numbers, and, Fermat-powers square to its sesquimultiple, that is 
. Moreover, all natural numbers smaller than 
form a finite field 
. Using distributivity, one can write down a multiplication table for all 2-powers.
The Knight-seating problems asks for a consistent placing of n-th Knight 
at an odd root of unity, compatible with the two different realizations of . Last time, we were able to place the first 15 Knights as below, and asked where you would seat 
was placed at 
as 4 was the smallest number generating the ‘Fermat’-field 
(with multiplicative group of order 15) subject to the compatibility relation with the generator 2 of the smaller Fermat-field (with group of order 15) that 
.
To include the next Fermat-field 
(with multiplicative group of order 255) consistently, we need to find the smallest number n generating the multiplicative group and satisfying the compatibility condition 
. Let’s first concentrate on finding the smallest generator : as 2 is a generator for 1st Fermat-field 
and 4 a generator for the 2-nd Fermat-field a natural conjecture might be that 16 is a generator for the 3-rd Fermat-field and, more generally, that 
would be a generator for the next field 
.
However, an “exercise” in the 1978-paper by Hendrik Lenstra Nim multiplication asks : “Prove that is a primitive root in the field if and only if i=0 or 1.”
I’ve struggled with several of the ‘exercises’ in Lenstra’s paper to the extend I feared Alzheimer was setting in, only to find out, after taking pen and paper and spending a considerable amount of time calculating, that they are indeed merely exercises, when looked at properly… (Spoiler-warning : stop reading now if you want to go through this exercise yourself).
In the picture above I’ve added in red the number 
to each of the involutions. Clearly, for each pair these numbers are all distinct and we see that for the indicated pairing they make up all numbers strictly less than 8.
By Conway’s simplicity rules (or by checking) the pair (16,17) gives the number 8. In other words, the equation
is an irreducible polynomial over 
having as its roots in 
the numbers 16 and 17. But then, 16 and 17 are conjugated under the Galois-involution (the Frobenius 
). That is, we have 
and 
and hence 
. Now, use the multiplication table in given in the previous post (or compute!) to see that 8 is of order 5 (and NOT a generator). As a consequence, the multiplicative order of 16 is 5×17=85 and so 16 cannot be a generator in .
For general i one uses the fact that and 
are the roots of the polynomial 
over and argues as before.
Right, but then what is the minimal generator satisfying ? By computing we see that the pairings of all numbers in the range 16…31 give us all numbers in the range 8…15 and by the above argument this implies that the 17-th powers of all numbers smaller than 32 must be different from 4. But then, the smallest candidate is 32 and one verifies that indeed 
(use the multiplication table given before).
Hence, we must place Knight 
at root 
and place the other Knights prior to the 256-th at the corresponding power of 32. I forgot the argument I used to find-by-hand the requested place for Knight 16, but one can verify that 
so we seat at root 
.
But what about Knight 
? Well, by this time I was quite good at squaring and binary representations of integers, but also rather tired, and decided to leave that task to the computer.
If we denote Nim-addition and multiplication by 
and 
, then Conway’s simplicity results in ONAG establish a field-isomorphism between 
and the field 
where the 
satisfy the Artin-Schreier equations
and the i-th Fermat-field corresponds to 
. The correspondence between numbers and elements from these fields is given by taking 
. But then, wecan write every 2-power as a product of the and use the binary representation of numbers to perform all Nim-calculations with numbers in these fields.
Therefore, a quick and dirty way (and by no means the most efficient) to do Nim-calculations in the next Fermat-field consisting of all numbers smaller than 65536, is to use sage and set up the field 
by
R.< x,y,z,t > =GF(2)[] S.< a,b,c,d >=R.quotient((x^2+x+1,y^2+y+x,z^2+z+x*y,t^2+t+x*y*z))
To find the smallest number generating the multiplicative group and satisfying the additional compatibility condition 
we have to find the smallest binary number 
(larger than 255) satisfying
(i1*a*b*c*t+i2*b*c*t+i3*a*c*t+i4*c*t+i5*a*b*t+i6*b*t+ i7*a*t+i8*t+i9*a*b*c+i10*b*c+i11*a*c+i12*c+i13*a*b+ i14*b+i15*a+i16)^257=a*c
It takes a 2.4GHz 2Gb-RAM MacBook not that long to decide that the requested generator is 1051 (killing another optimistic conjecture that these generators might be 2-powers). So, we seat Knight
at root 
and can then arrange seatings for all Knight queued up until we reach the 65536-th! In particular, the first Knight we couldn’t place before, that is Knight , will be seated at root 
.
If you’re lucky enough to own a computer with more RAM, or have the patience to make the search more efficient and get the seating arrangement for the next Fermat-field, please drop a comment.
I’ll leave you with another Lenstra-exercise which shouldn’t be too difficult for you to solve now : “Prove that 
has three solutions in 
for each 
.”
big Witt vectors for everyone (1/2)
Feb 2nd
Next time you visit your math-library, please have a look whether these books are still on the shelves : Michiel Hazewinkel‘s Formal groups and applications, William Fulton’s and Serge Lange’s Riemann-Roch algebra and Donald Knutson’s lambda-rings and the representation theory of the symmetric group.
I wouldn’t be surprised if one or more of these books are borrowed out, probably all of them to the same person. I’m afraid I’m that person in Antwerp…
Lately, there’s been a renewed interest in -rings and the endo-functor W assigning to a commutative algebra its ring of big Witt vectors, following Borger’s new proposal for a geometry over the absolute point.
However, as Hendrik Lenstra writes in his 2002 course-notes on the subject Construction of the ring of Witt vectors : “The literature on the functor W is in a somewhat unsatisfactory state: nobody seems to have any interest in Witt vectors beyond applying them for a purpose, and they are often treated in appendices to papers devoting to something else; also, the construction usually depends on a set of implicit or unintelligible formulae. Apparently, anybody who wishes to understand Witt vectors needs to construct them personally. That is what is now happening to myself.”
Before doing a series on Borger’s paper, we’d better run through Lenstra’s elegant construction in a couple of posts. Let A be a commutative ring and consider the multiplicative group of all ‘one-power series’ over it ![\Lambda(A)=1+t A[[t]]](/latexrender/pictures/52459f7e0a68f4e61fb742f8cb1a710e.gif "\Lambda(A)=1+t A[[t]]")
. Our aim is to define a commutative ring structure on taking as its ADDITION the MULTIPLICATION of power series.
That is, if 
, then we define our addition 
. This may be slightly confusing as the ZERO-element in 
will then turn be the constant power series 1…
We are now going to define a multiplication 
on which is distributively with respect to 
and turns into a commutative ring with ONE-element the series 
.
We will do this inductively, so consider 
the (classes of) one-power series truncated at term n, that is, the kernel of the natural augmentation map between the multiplicative group-units ![~A[t]/(t^{n+1})^* \rightarrow A^*](/latexrender/pictures/44e56a0ed489c1b14d5a7fd48d5b009f.gif "~A[t]/(t^{n+1})^* \rightarrow A^*")
.
Again, taking multiplication in ![A[t]/(t^{n+1})](/latexrender/pictures/069719475f6101cacac1eb7ca27a289b.gif "A[t]/(t^{n+1})")
as a new addition rule , we see that 
is an Abelian group, whence a 
-module.
For all elements we have a scaling operator 
(sending 
) which is an A-ring endomorphism of , in particular multiplicative wrt. 
. But then, is an additive endomorphism of , so is an element of the endomorphism-RING 
. Because composition (being the multiplication in this endomorphism ring) of scaling operators is clearly commutative (
) we can define a commutative RING 
being the subring of generated by the operators .
The action turns into an E-module and we define an E-module morphism 
by 
.
All of this looks pretty harmless, but the upshot is that we have now equipped the image of this E-module morphism, say 
(which is the additive subgroup of generated by the elements ) with a commutative multiplication induced by the rule .
Explicitly, is the set of one-truncated polynomials 
with coefficients in such that one can find elements 
such that 
. We multiply with another such truncated one-polynomial 
(taking elements 
) via
and using distributivity and the multiplication rule this gives the element 
.
Being a ring-qutient of we have that 
is a commutative ring, and, from the construction it is clear that 
behaves functorially.
For rings such that 
we are done, but in general may be strictly smaller. The idea is to use functoriality and do the relevant calculations in a larger ring 
where we can multiply the two truncated one-polynomials and observe that the resulting truncated polynomial still has all its coefficients in .
Here’s how we would do this over : take two irreducible one-polynomials u(t) and v(t) of degrees r resp. s smaller or equal to n. Then over the complex numbers we have
and 
. Then, over the field 
we have that 
and hence we can compute their product 
as before to be 
. But then, all coefficients of this truncated K-polynomial are invariant under all permutations of the roots 
and the roots 
and so is invariant under all elements of the Galois group. But then, these coefficients are algebraic numbers in 
whence integers. That is, 
. It should already be clear from this that the rings 
contain a lot of arithmetic information!
For a general commutative ring we will copy this argument by considering a free overring 
(with 1 as one of the base elements) by formally adjoining roots. At level 1, consider 
to be the set of all non-constant one-polynomials over and consider the ring
![A^{(1)} = \bigotimes_{f \in M_0} A[X]/(f) = A[X_f, f \in M_0]/(f(X_f) , f \in M_0)](/latexrender/pictures/c6e929599a48704975bfa805ec79901d.gif "A^{(1)} = \bigotimes_{f \in M_0} A[X]/(f) = A[X_f, f \in M_0]/(f(X_f) , f \in M_0)")
The idea being that every one-polynomial 
now has one root, namely 
in 
. Further, is a free A-module with basis elements all 
with 
.
Good! We now have at least one root, but we can continue this process. At level 2, 
will be the set of all non-constant one-polynomials over and we use them to construct the free overring 
(which now has the property that every has at least two roots in ). And, again, we repeat this process and obtain in succession the rings 
. Finally, we define 
having the property that every one-polynomial over A splits entirely in linear factors over .
But then, for all 
we can compute 
. Remains to show that the resulting truncated one-polynomial has all its entries in A. The ring 
contains two copies of namely 
and 
and the intersection of these two rings in exactly (here we use the freeness property and the additional fact that 1 is one of the base elements). But then, by functoriality of , the element
lies in the intersection 
. Done!
Hence, we have endo-functors 
in the category of all commutative rings, for every number n. Reviewing the construction of one observes that there are natural transformations 
and therefore also natural transformations 
. Taking the inverse limits 
we therefore have the ‘one-power series’ endo-functor
which is ‘almost’ the functor W of big Witt vectors. Next time we’ll take you through the identification using ‘ghost variables’ and how the functor 
can be used to define the category of -rings.
The odd knights of the round table
Jan 28th
Here’s a tiny problem illustrating our limited knowledge of finite fields : “Imagine an infinite queue of Knights 
, waiting to be seated at the unit-circular table. The master of ceremony (that is, you) must give Knights 
and 
a place at an odd root of unity, say 
and 
, such that the seat at the odd root of unity 
must be given to the Knight 
, where 
is the Nim-multiplication of 
and 
. Which place would you offer to Knight , or Knight , or, if you’re into ordinals, Knight 
?”
What does this have to do with finite fields? Well, consider the simplest of all finite field 
and consider its algebraic closure . Last year, we’ve run a series starting here, identifying the field , following John H. Conway in ONAG, with the set of all ordinals smaller than , given the Nim addition and multiplication. I know that ordinal numbers may be intimidating at first, so let’s just restrict to ordinary natural numbers for now. The Nim-addition of two numbers 
can be calculated by writing the numbers n and m in binary form and add them without carrying. For example, 
. Nim-multiplication is slightly more complicated and is best expressed using the so-called Fermat-powers 
. We then demand that 
whenever 
and 
. Distributivity wrt. can then be used to calculate arbitrary Nim-products. For example, 
. Conway’s remarkable result asserts that the ordinal numbers, equipped with Nim addition and multiplication, form an algebraically closed field of characteristic two. The closure is identified with the subfield of all ordinals smaller than . For those of you who don’t feel like going transfinite, the subfield is identified with the quadratic closure of .
The connection between and the odd roots of unity has been advocated by Alain Connes in his talk before a general public at the IHES : “L’ange de la géométrie, le diable de l’algèbre et le corps à un élément” (the angel of geometry, the devil of algebra and the field with one element). He describes its content briefly in this YouTube-video
At first it was unclear to me which ‘coupling-problem’ Alain meant, but this has been clarified in his paper together with Caterina Consani Characteristic one, entropy and the absolute point. The non-zero elements of can be identified with the set of all odd roots of unity. For, if x is such a unit, it belongs to a finite subfield of the form 
for some n, and, as the group of units of any finite field is cyclic, x is an element of order 
. Hence, 
can be identified with the set of -roots of unity, with 
corresponding to a generator of the unit-group. So, all elements of correspond to an odd root of unity. The observation that we get indeed all odd roots of unity may take you a couple of seconds1.
Assuming we succeed in fixing a one-to-one correspondence between the non-zero elements of and the odd roots of unity 
respecting multiplication, how can we recover the addition on ? Well, here’s Alain’s coupling function, he ties up an element x of the algebraic closure to the element s(x)=x+1 (and as we are in characteristic two, this is an involution, so also the element tied up to x+1 is s(x+1)=(x+1)+1=x. The clue being that multiplication together with the coupling map s allows us to compute any sum of two elements as 
.
For example, all information about the finite field 
is encoded in this identification with the 15-th roots of unity, together with the pairing s depicted as
Okay, we now have two identifications of the algebraic closure : the smaller ordinals equipped with Nim addition and Nim multiplication and the odd roots of unity with complex-multiplication and the Connes-coupling s. The question we started from asks for a general recipe to identify these two approaches.
To those of you who are convinced that finite fields (LOL, even characteristic two!) are objects far too trivial to bother thinking about : as far as I know, NOBODY knows how to do this explicitly, even restricting the ordinals to merely the natural numbers!
Please feel challenged! To get you started, I’ll show you how to place the first 15 Knights and give you a procedure (though far from explicit) to continue. Here’s the Nim-picture compatible with that above
To verify this, and to illustrate the general strategy, I’d better hand you the Nim-tables of the first 16 numbers. Here they are
It is known that the finite subfields of are precisely the sets of numbers smaller than the Fermat-powers 
. So, the first one is all numbers smaller than 
(check!). The smallest generator of the multiplicative group (of order 3) is 2, so we take this to correspond to the unit-root 
. The next subfield are all numbers smaller than 
and its multiplicative group has order 15. Now, choose the smallest integer k which generates this group, compatible with the condition that 
. Verify that this number is 4 and that this forces the identification and coupling given above.
The next finite subfield would consist of all natural numbers smaller than 
. Hence, in this field we are looking for the smallest number k generating the multiplicative group of order 255 satisfying the extra condition that 
which would fix an identification at that level. Then, the next level would be all numbers smaller than 
and again we would like to find the smallest number generating the multiplicative group and such that the appropriate power is equal to the aforementioned k, etc. etc.
Can you give explicit (even inductive) formulae to achieve this? I guess even the problem of placing Knight 16 will give you a couple of hours to think about… (to be continued).
- If m is odd, then (2,m)=1 and so 2 is a unit in the finite cyclic group
whence 
, so the m-roots of unity lie within those of order [↩]
Hi, I'm a professor of mathematics at the University of Antwerp, in Belgium. My research areas are non-commutative algebra and non-commutative geometry. Sometimes, you can also find me here :
